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the speed of a car increases from 10m/s to 20m/s with an acceleration of 3.0 m/s^2. Find the distance covered during the acceleration.
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v^=u^2+2as 20x20=10x10+2x3xs s=distance covered 400=100+6s s=50meters
by 3rd eqn of motion
I got it now. thanks a lot :)
This is again conservation of energy. Work = (Force)x(distance) = Fs = mas But Work = Change in KE = (1/2)mv^2 - (1/2)mu^2 = (m/2)(v^2 - u^2) Hence mas = (m/2)(v^2 - u^2) or v^2 - u^2 = 2as
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